# special values of symmetric functions at powers of $\frac1j$

Let $$e_n(x_1,x_2,x_3,\dots)$$ denote the $$n$$-th elementary symmetric function in the infinite variables $$x_1,x_2,x_3,\dots$$.

Let $$u$$ and $$v$$ be the roots of $$z^2-6z+1=0$$.

Question. Let $$x_j=\frac1{j^8}$$. The following seems to be true, but can one prove or disprove? $$e_n(x_1,x_2,x_3,\dots)=\frac{4^{3n+1}\pi^{8n}(u^{2n+1}+v^{2n+1})}{(8n+4)!}.$$

For example, $$e_0=1$$ and $$e_1=\sum_{j\geq1}\frac1{j^8}=\zeta(8)=\frac{\pi^8}{9450}$$.

• In other words, you're asking for the evaluation of the infinite product $\prod_{j=1}^{+\infty} (\frac{1}{j^8}-t)$, right? – Gro-Tsen Mar 24 '17 at 10:58
• That's is correct. – T. Amdeberhan Mar 24 '17 at 11:03

As Gro-Tsen suggests in the comments, we have to expand the infinite product $$f(t)=\prod_j \left(1-\frac{t^8}{j^8}\right)=\prod_{j;\,w^4=1} \left(1-\frac{\omega t^2}{j^2}\right)=\prod_{w^4=1}\frac{\sin\pi\sqrt{w}t}{\pi\sqrt{w}t},$$ we expand product of four sines as an alternating sum of cosines $$\sin a\sin b\sin c\sin d=\frac18\sum (\prod \pm)\cdot\cos(a\pm b\pm c\pm d),$$ and write Taylor series for cosines. Powers of $1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)$ appear. To be more concrete, $c_n:=e_n(x_1,\dots)$ equals $(-1)^n\times [t^{8n}]f(t)$, where $[t^n]g$ denotes a coefficient of $t^n$ in $g$. Thus $$c_n=(-1)^{n+1}i\pi^{8n}\frac1{8(8n+4)!}\sum \pm\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^{8n+4}.$$ We have $$\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^4=8i(\pm 3\pm 2\sqrt{2}),$$ that gives your formula after careful simplifications.